# A proton with velocity v completes a circle with radius r in uniform magnetic field.what should be the velocity of alpha particle so that it completes a circle with same radius under the influence of the same magnetic field?

Aug 8, 2018

The force experienced by a particle of charge $q$ moving in a magnetic field perpendicular to the direction of the field acts perpendicularly to both velocity and magnetic field and provides the centrpetal force required for the motion of the particle in a circular path.For the dynamic equilibrium established here we have the following relation

$\frac{m {v}^{2}}{r} = q v {B}_{m}$

$\implies v = \frac{q r {B}_{m}}{m}$

Where

$m \to \text{mass of the particle}$

$v \to \text{speed of the particle}$

$r \to \text{radius of circular path}$

${B}_{m} \to \text{intensity of magnetic field}$

In our problem both proton and $\alpha$ particle follow circular path of same radius under the influence of the same magnetic field.

For proton the above relation becomes

$\implies v = \frac{{q}_{p} r {B}_{m}}{m} _ p$

And for $\alpha$ particle the above relation becomes

$\implies {v}_{\alpha} = \frac{{q}_{\alpha} r {B}_{m}}{m} _ \alpha$

So ${v}_{\alpha} / v = \frac{{m}_{p} {q}_{\alpha}}{{m}_{\alpha} {q}_{p}} = \frac{1}{4} \cdot \frac{2}{1} = \frac{1}{2}$

Hence ${v}_{\alpha} = \frac{v}{2}$